TITLE: Does Multiplicative Version of Azuma's Inequality Hold?
QUESTION [8 upvotes]: It is known that there are multiplicative version concentration inequalities for
sums of independent random variables. For example, the following
multiplicative version Chernoff bound.

Chernoff bound:
Let $X_1,\ldots,X_n$ be independent random variables and $X_i \in$
$[0,1]$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$,
$\Pr\left(Y \ge (1+\delta)EY \right) \le e^{-c\cdot(EY)\delta ^2},$
where $c$ is some absolute constant, e.g., c=1/3.

Now consider dependent random variables. A slight variant of Azuma's inequality states the following.

Azuma's Inequality:
Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in
[0,1]$. Assume that there exists $\mu$, such that $ \Pr \left( \sum_{i=1}^n \mathbb{E}[X_i|X_{1},\ldots,X_{i-1}] \le \mu\right) = 1$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\lambda > 0$,
$\Pr\left(Y \ge n\mu+\lambda \right) \le e^{-2 \lambda^2/n}.$

Azuma's inequality is additive. My question is that does a
multiplicative version of Azuma's inequality such as the following
hold?

My question: does the following hold?
Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in
[0,1]$. Assume that there exists $\mu$, such that $\Pr\left( \sum_{i=1}^n \mathbb{E}[X_i|X_1,\ldots,X_{i-1}] \le \mu\right) = 1.$ Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$
$\Pr\left(Y \ge (1+\delta)n\mu \right) \le e^{-c\cdot n\mu \delta^2},$
where $c$ is some absolute constant.

Note: the standard Azuma's inequality does not imply the
multiplicative version when $n\mu \ll
\sqrt{n}$.

REPLY [4 votes]: $\newcommand{\de}{\delta}$
The "dependent" version of the multiplicative Chernoff bound can be proved quite similarly to the "independent" case. Indeed, let $E_{i-1}$ denote the conditional expectation given $X_1,\dots,X_{i-1}$, so that $E_{i-1}X_i\le\mu$ almost surely (a.s.) for $i=1,\dots,n$. Take any real $t\ge0$. By the convexity of $e^{tx}$ in $x$ and the conditions that $0\le X_i\le1$ and $E_{i-1}X_i\le\mu$, we have $e^{tX_i}\le1+(e^t-1)X_i$ and hence
\begin{equation*}
 E_{i-1}e^{tX_i}\le1+(e^t-1)E_{i-1}X_i\le1+(e^t-1)\mu\le\exp\{(e^t-1)\mu\} 
\end{equation*}
for $i=1,\dots,n$. So, by induction, for $j=1,\dots,n$ and $Y_j:=\sum_1^j X_i$ we have 
\begin{equation*}
 Ee^{tY_j}=EE_{j-1}e^{tY_j}=Ee^{tY_{j-1}}E_{j-1}e^{tX_j}\le Ee^{tY_{j-1}}\exp\{(e^t-1)\mu\},
\end{equation*}
whence, by induction, 
\begin{equation*}
 Ee^{tY}=Ee^{tY_n}\le\exp\{n(e^t-1)\mu\}. 
\end{equation*}
So, using Markov's inequality and then choosing $t=\ln(1+\de)$, we get 
\begin{align*}
 P(Y\ge(1+\de)n\mu)&\le e^{-(1+\de)n\mu t}Ee^{tY}
 \le\exp\{-(1+\de)n\mu t+n(e^t-1)\mu\} \\ 
 &=\exp\{-n\mu\psi(\de)\}, 
\end{align*}
where $\psi(u):=u-(1+u)\ln(1+u)$. 
Up to notation, this bound is the same as the known multiplicative Chernoff bound in the "independent" case.
Since $\psi(u)\le-u^2/3$ for $u\in[0,3/2]$, we have 
\begin{equation*}
 P(Y\ge(1+\de)n\mu)\le\exp\{-n\mu\de^2/3\} \tag{1}
\end{equation*}
if $\de\in[0,3/2]$. 
Note that (1) cannot hold for all $\de\ge0$, even in the "independent" case. Indeed, suppose that the $X_i$'s are iid with $P(X_1=1)=\mu=1-P(X_i=0)$, where $\mu:=1/n$ and $n\to\infty$. Then $Y$ will converge in distribution to  a random variable $\Pi$ with the Poisson distribution with parameter $1$, and (1) will yield 
\begin{equation*}
 P(\Pi\ge1+\de)\le\exp\{-\de^2/3\}.  
\end{equation*}
Since Poisson distributions are not subgaussian, the latter inequality cannot hold for all $\de\ge0$. So, (1) cannot hold for all $\de\ge0$.